Divide the following complex numbers. $ \dfrac{-9-19i}{3-5i}$
We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${3+5i}$ $ \dfrac{-9-19i}{3-5i} = \dfrac{-9-19i}{3-5i} \cdot \dfrac{{3+5i}}{{3+5i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-9-19i) \cdot (3+5i)} {(3-5i) \cdot (3+5i)} = \dfrac{(-9-19i) \cdot (3+5i)} {3^2 - (-5i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-9-19i) \cdot (3+5i)} {(3)^2 - (-5i)^2} = $ $ \dfrac{(-9-19i) \cdot (3+5i)} {9 + 25} = $ $ \dfrac{(-9-19i) \cdot (3+5i)} {34} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-9-19i}) \cdot ({3+5i})} {34} = $ $ \dfrac{{-9} \cdot {3} + {-19} \cdot {3 i} + {-9} \cdot {5 i} + {-19} \cdot {5 i^2}} {34} $ Evaluate each product of two numbers. $ \dfrac{-27 - 57i - 45i - 95 i^2} {34} $ Finally, simplify the fraction. $ \dfrac{-27 - 57i - 45i + 95} {34} = \dfrac{68 - 102i} {34} = 2-3i $